What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point
1 answer:
<u>Answer:</u> The amount of energy absorbed by water is 5390 Calories
<u>Explanation:</u>
To calculate the amount of heat absorbed at normal boiling point, we use the equation:
where,
q = amount of heat absorbed = ?
m = mass of water = 10 grams
= latent heat of vaporization = 539 Cal/g
Putting values in above equation, we get:
Hence, the amount of energy absorbed by water is 5390 Calories
You might be interested in
Answer:
The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,
Explanation:
To answer this problem we need to keep in mind the <u>definition of pH</u>:
- pH = -log [H⁺]
So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:
- 5.2 = -log [H⁺]
-5.2 = log [H⁺]
= [H⁺]
6.31 x 10⁻⁶ M = [H⁺]
- 5.6 = -log [H⁺]
-5.6 = log [H⁺]
= [H⁺]
2.51 x 10⁻⁶ M = [H⁺]